Projectile Motion: Calculating Velocity After 2 Seconds
Alright guys, let's dive into a classic physics problem! We're going to figure out the velocity of a body in projectile motion after a certain amount of time. This is super important for understanding how things move when they're launched into the air, like a ball you throw or a rocket taking off. So, let's break down the problem and get to the solution. We are going to be talking about projectile motion, velocity, time, angle, and maximum height.
The Problem: Setting the Stage
So, here's the scenario, and pay close attention because we're going to need these details to solve the problem. A body is launched from the ground. It's fired at an angle, which we'll call θ, relative to the horizontal. The initial velocity of this body is 30 m/s (meters per second). Now, as it flies through the air, gravity is working on it, pulling it down. Because of gravity, the body follows a curved path, a parabola to be exact, until it hits the ground again. The body reaches a maximum height of 11.25 m. Our main goal here is to calculate the velocity of the body 2 seconds after it's launched. Understanding this allows us to predict where the object will land and its velocity at any given point. It's all about applying the right physics principles and a little bit of math. The good news is that once you grasp the core concepts, these problems become much easier. Are you ready to solve it? Let's go.
First, we need to brush up on a few key concepts. Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. The object moves in two dimensions: horizontally and vertically. The horizontal component of the velocity remains constant (assuming we ignore air resistance), while the vertical component changes due to gravity. Maximum height is the highest point the projectile reaches in its trajectory. At this point, the vertical component of the velocity is zero. We'll also need to understand how to break down the initial velocity into its horizontal and vertical components using trigonometry. We can use this information to understand how far the projectile will travel horizontally (range), how high it will go (maximum height), and how long it will stay in the air (time of flight). Projectile motion problems often seem complicated at first, but breaking them down step by step makes them much more manageable. Let's tackle this problem with a clear plan of attack. We'll use the given information to find the initial launch angle. Then, we'll figure out the horizontal and vertical components of the initial velocity. Finally, we'll use these components and the time (2 seconds) to calculate the final velocity. Remember, understanding the basics is key to solving more complex physics problems. Always take it one step at a time and ask questions when something doesn't make sense. Are you with me?
Unpacking the Concepts: Projectile Motion Fundamentals
Before we jump into the calculations, let's quickly recap some important concepts. Projectile motion, as we mentioned, is when an object moves through the air, and the only force acting on it is gravity. Now, gravity pulls things downwards, which affects the vertical component of the projectile's velocity. The horizontal component, however, remains constant because we're (usually) ignoring air resistance. Think of it like this: If you throw a ball, it moves forward horizontally, but gravity constantly pulls it down, making it follow a curved path. The initial velocity is the speed and direction with which the object is launched. This velocity can be broken down into two components: horizontal and vertical. The angle at which the object is launched (the launch angle) determines the proportions of the horizontal and vertical components. The maximum height is the highest point the object reaches in its flight. At this point, the vertical velocity is zero, but the horizontal velocity remains constant. The time it takes for the object to reach its maximum height and double that time is the time of flight.
We can calculate the horizontal range (the distance the object travels horizontally) using the initial velocity, launch angle, and time of flight. Understanding these fundamental concepts is crucial for solving any projectile motion problem. Now let's make sure we're all on the same page. The following variables and equations are going to be used:
- v₀: Initial velocity (30 m/s)
- θ: Launch angle (to be determined)
- h: Maximum height (11.25 m)
- t: Time (2 s)
- g: Acceleration due to gravity (approximately 9.8 m/s²)
Knowing these allows us to solve the problem step by step.
Step 1: Finding the Launch Angle (θ)
Alright, the first step is to figure out the launch angle, θ. We know the maximum height (h) and the initial velocity (v₀), so we can use a formula to find θ. The formula we'll use is derived from the principles of projectile motion and relates the maximum height, initial velocity, and launch angle. The formula is:
h = (v₀² sin² θ) / (2g)
Where:
- h is the maximum height
- v₀ is the initial velocity
- g is the acceleration due to gravity (9.8 m/s²)
Let's rearrange this formula to solve for θ. First, multiply both sides by 2g:
2g h = v₀² sin² θ
Then, divide both sides by v₀²:
(2g h) / v₀² = sin² θ
Now, take the square root of both sides to solve for sin θ:
sin θ = √((2g h) / v₀²)
Finally, take the inverse sine (sin⁻¹) of both sides to find θ:
θ = sin⁻¹(√((2g h) / v₀²))
Now, plug in the numbers:
θ = sin⁻¹(√((2 * 9.8 m/s² * 11.25 m) / (30 m/s)²))
θ = sin⁻¹(√(220.5 / 900))
θ = sin⁻¹(√0.245)
θ = sin⁻¹(0.495)
θ ≈ 29.67°
So, the launch angle θ is approximately 29.67 degrees. This is super important because it tells us the direction the body was initially launched. Understanding the launch angle is key to analyzing the projectile's trajectory. Now that we know the launch angle, we can move on to the next step.
Step 2: Calculate Initial Velocity Components
In projectile motion, the initial velocity is broken down into two components: horizontal (x) and vertical (y). Now that we know the launch angle (θ), we can calculate these components using trigonometry. The horizontal component (v₀x) is given by:
v₀x = v₀ cos θ
The vertical component (v₀y) is given by:
v₀y = v₀ sin θ
Let's plug in the values:
v₀x = 30 m/s * cos(29.67°)
v₀x ≈ 30 m/s * 0.869
v₀x ≈ 26.07 m/s
v₀y = 30 m/s * sin(29.67°)
v₀y ≈ 30 m/s * 0.495
v₀y ≈ 14.85 m/s
So, the initial horizontal velocity (v₀x) is approximately 26.07 m/s, and the initial vertical velocity (v₀y) is approximately 14.85 m/s. These components are super helpful because they allow us to track the motion of the body separately in the horizontal and vertical directions. The horizontal velocity remains constant, while the vertical velocity changes due to gravity. Knowing these components helps us analyze the entire trajectory of the object.
Step 3: Determine the Velocity After 2 Seconds
Now for the grand finale, let's find the velocity of the body after 2 seconds. Remember, the horizontal velocity (vx) remains constant throughout the motion because we're ignoring air resistance. So, vx = v₀x ≈ 26.07 m/s. To find the vertical velocity (vy) after 2 seconds, we use the following equation:
vy = v₀y - g t
Where:
- v₀y is the initial vertical velocity (14.85 m/s)
- g is the acceleration due to gravity (9.8 m/s²)
- t is the time (2 s)
Plugging in the values:
vy = 14.85 m/s - (9.8 m/s² * 2 s)
vy = 14.85 m/s - 19.6 m/s
vy ≈ -4.75 m/s
The negative sign indicates that the vertical velocity is directed downwards. Now that we have vx and vy at t = 2 s, we can calculate the magnitude and direction of the resultant velocity. The magnitude of the velocity (v) is given by:
v = √(vx² + vy²)
v = √(26.07² + (-4.75)²)
v = √(679.64 + 22.56)
v = √702.2
v ≈ 26.5 m/s
The direction of the velocity can be found using the arctangent function:
α = tan⁻¹(vy / vx)
α = tan⁻¹(-4.75 / 26.07)
α ≈ -10.3°
Therefore, the velocity of the body after 2 seconds is approximately 26.5 m/s at an angle of -10.3 degrees (below the horizontal). Excellent work, guys! This means that after two seconds, the body is moving at approximately 26.5 m/s and is slightly angled downwards. This detailed breakdown helps in understanding the nuances of projectile motion and the effects of gravity on an object's trajectory. This result confirms how the initial conditions (velocity and angle) influence the projectile's future positions.
Conclusion: Putting It All Together
So, to recap, we've solved a projectile motion problem. We started with the initial velocity and the maximum height, and from that, we calculated the launch angle. We then found the horizontal and vertical components of the initial velocity. Finally, we calculated the velocity of the body 2 seconds after it was launched. The final velocity was approximately 26.5 m/s at an angle of about -10.3 degrees. This problem illustrates the importance of understanding the components of projectile motion and how they change over time. By breaking down the problem into manageable steps and using the right formulas, we were able to find the solution. Awesome job! This is a great foundation for tackling even more complex physics problems. Keep practicing, and you'll become a pro in no time. Always remember the basics: The horizontal velocity is constant, and the vertical velocity is affected by gravity. Also, don't be afraid to draw diagrams and break the problem into smaller parts. And most importantly, practice makes perfect! Keep up the great work, and you'll master physics in no time! Feel free to ask if you have any questions or want to dive deeper into any of the concepts we covered today. Keep learning, keep exploring, and always stay curious!
